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Let $G$ be a finite group with faithful irreducible representation $\gamma: G \to GL_n(\mathbb{C})$, $n>1$.

Can we put a bound on the size of $G$? What if $G$ is nilpotent?

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Every finite cyclic group has a 1-dimensional faithful representation... – Alain Valette Oct 2 '12 at 21:27
I remembered this point when I was thinking about asking a quesiton, but I forgot to put this in the question! The question is now adjusted to eliminate the cyclic case. Thanks for the reminder though Alain! – Alonso Oct 2 '12 at 21:33
$\mathbf{C}[G]$ is ths sum of $G$'s irreps, so certainly you deduce that $n\le |G|$, which can be improved. But if you want an upper bound I don't think you can say much just by artificially removing $n=1$. – YCor Oct 2 '12 at 22:37
Every dihedral group of order 6 or more has a faithful irreducible representation of degree 2. Which also answers the nilpotent case if you take dihedral groups of order a power of 2. – Jeremy Rickard Oct 2 '12 at 23:00
If you like, you can think of Alain's comment as: large subgroups of roots of unity in $\mathbb{C}^∗=GL_1(\mathbb{C})$. – LMN Oct 2 '12 at 23:28
up vote 11 down vote accepted

Jordan's theorem says that there is a function $f : \mathbb{N} \to \mathbb{N}$ such that whenever $G$ is a finite subgroup of ${\rm GL}(n,\mathbb{C}),$ there is an Abelian onrmal subgroup $A$ such that $[G:A] \leq f(n)$. Explicit bounds were given later, which can be much improved by invoking the classification of finite simple groups. If $G$ is a primitive subgroup of ${\rm SL}(n,\mathbb{C})$, then it follows from Jordan's theorem that $|G|$ is bounded in terms of $n$. Recall that a primitive representation is an irreducible one which is not equivalent to a representation induced from a representation of a proper subgroup.

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Did Jordan used modern language? Just curious.... – Alexander Chervov Oct 3 '12 at 3:55
I am not sure, but the proofs by Frobenius, Schur and Burnside which give explicit bounds use fairly modern language. Once you know that your group may be assumed to consist of unitary matrices, Jordan's theorem is essentially a compactness argument- it can be reduced to the primitive case, and then it follows that if $x$ and $y$ are in diffent cosets of $Z(G)$, then we have $\|x - y \| \geq \frac{1}{2}$ where we use the operator norm on $M_{n}(\mathbb{C}).$ Since $G$ is contained in the ball of $M_{n}(\mathbb{C})$ (even in its boundary), we are done. – Geoff Robinson Oct 3 '12 at 18:23

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